This year, Squares and Factorials

Claim: Only one pair of positive integers $(x,y)$ satisfy the given equation, i.e $\boxed{(x,y)=(45, 3)}$

Proof: Note that $4$ divides $y!$ for all $y\geq 4.$ So for all $y\geq 4$ , $x^2 -y!\equiv x^2\equiv 2019\equiv 3 \pmod{4}$ But $3$ is not a quadratic residue modulo $4$ . (Squares leave the remainder $0$ or $1$ when divided by $4$ .) Hence the only cases left to check are $y=1, 2, 3.$

The only case that gives a solution is $y=3\implies x=45$ .

Relevant wiki: Modular Arithmetic and Quadratic Residues

If $x^2 +y! =2019$ , then $2019+y!=x^2$

• if y = 1,2 , we have no solution since 2020 and 2021 aren’t squares
• if y>2, y! is divisible by 3, as is 2019
• square numbers have even amounts of prime factors, so if 3’s a factor, 9 must be $(y!+2019)/3 = \frac{y!}{3} + 673$ must be divisible by 3
• If $y\geq 6$ , y!/3 is divisible by 3, so the above condition cannot be satisfied
• of 3,4,5, only 3 gives an integer result for x, so the answer is $\boxed{1}$